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There are $n$ questions in an examination ($n \in \mathbb{N}$), and the answer to each question is True or False. You know that exactly $t$ of the answers are True ($0 \le t \le n$), so you randomly answer $t$ questions as True and the rest as False. What is your probability of getting at least a $50\%$ score in the exam (in terms of $n$ and $t$)?

My attempt:

WLOG assume you answer the first $t$ questions as True. Let there be $k$ questions out of the first $t$ of which the answer is True. Thus, out of the other $n-t$ questions where you replied False, $(n-t)-(t-k)=n-2t+k$ questions are really False. Thus, the fraction of correct answers for the whole examination is equal to $\frac{n-2t+2k}{n}$. If this is at least $1/2$, then $n+2k \ge 4t$. However, I can't calculate the probability of this happening.

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  • $\begingroup$ A suggestion: Even-case: The probability to answer at least $t/2$ of the answers right is $\sum\limits_{k=\frac{t}{2}}^n \binom{n}{k}\cdot \left(\frac12\right)^n$ $\endgroup$ Dec 3, 2021 at 14:45
  • $\begingroup$ If there are $10$ questions and $9$ of them are true and $1$ false and you answer $9$ as true, you are certain to get more than $50 \%$ score. So $t / n$ matters. $\endgroup$
    – Math Lover
    Dec 3, 2021 at 14:52

1 Answer 1

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There is a typo. Instead of $n+2k \ge 4t$, it should be $n+4k \ge 4t$. I do not know if you can get a closed form but here is my work that may help you understand lower and upper bound of $t$ in terms of $n$, beyond which one is certain to get at least $50\%$ score.

a) Based on your work, $ \displaystyle k \geq \frac{4t - n}{4}$. So if $t \geq \dfrac{3n}{4}, k \geq \dfrac{n}{2}$ and we are bound to get at least $50\%$ score by choosing $t$ answers as TRUE.

Similarly if $ ~\displaystyle t \leq \frac{n}{4}$, you are certain to get at least $50\%$ score by randomly choosing $(n-t)$ answers as FALSE.

b) Now for $~ \displaystyle \frac n4 \lt t \lt \frac {3n}4$,

we must have $ \displaystyle \lceil \frac{4t - n}{4} \rceil \leq k \leq t$ to score at least $50\%$

So the desired probability can be written as,

$ \displaystyle \sum \limits_{k = k_l}^t {t \choose k} {n - t \choose t - k} / {n \choose t}$

where $ \displaystyle k_l = \lceil \frac{4t - n}{4} \rceil$

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